How do you solve $2(x+4)^(2/3) = 8$ ?

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Answer 1 · 2015-04-09T17:33:11

Note that $q^(2/3) = (q^2)^(1/3)$

So
$2(x+4)^(2/3) = 8$

can be evaluated as
$2(x^2+8x+16)^(1/3) = 8$

cube both sides giving:
$8 (x^2+8x+16) = 8xx8xx8$
or
$x^2+8x+16 = 64$
$rarr$
$x^2+8x-48=0$

factoring:
$(x+12)(x-4) = 0$

So
$x=-12$
or
$x=4$

How do you solve 2(x+4)^(2/3) = 82(x+4)^(2/3) = 8?